# How could I prove that $x^2+1$ and $x^4+1$ are irreductible over $\mathbb{Z}/2 \mathbb{Z}$ and $\mathbb{Z}/3 \mathbb{Z}$? [duplicate]

How could I prove that $x^2+1$ and $x^4+1$ are irreductible over $\mathbb{Z}/2 \mathbb{Z}$ and $\mathbb{Z}/3 \mathbb{Z}$?

Theorem : Let $A$ an integral domain and $I$ a proper ideal of $A$. If $f(x) \not \equiv a(x)b(x) (\mod I)$ for any polynomials $a(x)$, $b(x)$ $\in A[x]$ of degree $\in [1, \deg(f))$, then $f(x)$ is irreductible in $A[x]$

I think I have to use this theorem, but I am not certain. Is anyone could help me at this point?

• Maybe I'm crazy, but I think these are reducible, at least in $\mathbb Z/2 \mathbb Z$. You can factor them into linear terms. – Alfred Yerger Mar 9 '16 at 3:33
• They first is not irreducible over $\mathbb Z/2\mathbb Z$ – Thomas Andrews Mar 9 '16 at 3:33
• Neither of those polynomials are irreducible in $\mathbb Z/2\mathbb 2Z$. A good first step before checking irreducibility is to see whether the polynomials have roots! In your case there are only finitely ($2$ and $3$ respectively) many candidates to check. – RKD Mar 9 '16 at 3:33
• What about $\mathbb{Z}/3\mathbb{Z}$? I would like to factorise $x^8-1$ in product of irreductibles in the rings $\mathbb{Z}/2\mathbb{Z}$ and $\mathbb{Z}/3\mathbb{Z}$? The problem is I really don't how to start. – Taj Mohamed Bandalandabad Mar 9 '16 at 3:36
• If you don't know how to start you can SEARCH for similar questions. Either here or in your lecture notes/textbook. You SHOULD NOT ASK THE SAME QUESTION MANY TIMES: – Jyrki Lahtonen Mar 9 '16 at 8:41

.$$x^2+1=x^2+1^2=(x+1)^2$$ in $$2\mathbb{Z}$$
$$x^4+1 = (x^2+x+2)(x^2+2x+2)$$ in $$3\mathbb{Z}$$
Neither is irreducible mod it's respective field. In fact, $$x^4+1$$ is reducible mod $$n\mathbb{Z}$$ for every $$n$$!